Multiplying out a three term polynomial; so in this particular example we are going to be looking at (4x plus y minus 7)². What this basically means is we are going to have, obviously the squared that the thing we have is going to be multiplied by itself.

What we could do is take each of these terms and distribute to each of the other terms and go through this process of having to multiply a bunch of things out, add up all our answers and combine like terms. But what I want to talk about is a way we can make our life easier, and what I would do is group a couple of items together. We have the same exact thing on either side, and so what we want to do is group two things. We can either group the 4x plus y or the y minus 7. It doesn’t matter which, okay?

So I’m going to do the y minus 7. We could just as easily have done the 4x plus y, either way we want to group two things together. From here we can make an easier substitution. So I say a is equal to 4x and b is equal to y minus 7, okay? A and B are arbitrary letters, it doesn’t matter you can choose L and K, whatever you want. I would avoid using X and Y because they are already represented in the problem, but any other letters should be fine.

So rewriting our problem with these two new terms. This term becomes a plus b, as does this one, so now we have a plus b times a plus b. We’ve taken our three term polynomial, turned them into two, okay? Using our laws of multiplication, FOILING this out, this ends up being a² plus 2ab plus b². And then doing some back substitution we know that a is actually 4x, so this becomes 4x². Our 2ab becomes +2, a is 4x and B is y minus 7 and +b² is equal to y minus 7 quantity squared.

Okay, so what we did is instead of having to distribute all this out we’ve turned it into a binomial and we expanded it up. How specific your teacher wants you to get can vary, some teachers will be perfectly happy with this, some are going to want you to distribute this out and simplify.

How we would do that, you know what 4x², here we have to distribute this through, we’ll have to multiply this one out. But each of those is significantly easier than having to take each of these terms and distribute it through.